Notes for Elementary Statistics User:Joel Vannatta Chapter 5
'Key terms' Random variable: A variable whose value is determined by the outcome of a random experiment Discrete random variable: A random variable which can be counted in whole numbers (i.e. number of cars) Continuous random variable: A random variable which can be measured at any amount within a range (i.e. height) Probability distribution: A chart which shows the different values of a discrete random variable alongside its frequency and/or relative frequency Example: In a probability distribution: #Relative frequency is always greater than 0 and less than 1 #The sum of all relative frequencies in a probability distribution is equal to 1 'Mean of a Discrete Random Variable' The mean of a discrete random variable is the value that is expected to occur, on average, if the experiment is repeated a large number of times. The equation for the mean of a discrete random variable is: μ = ΣxP(x), where: μ = mean x = random variable P(x) probability of random variable Example: ΣxP(x) = 0 + .3 + .7 + .3 = 1.3 μ = 1.3 'Standard Deviation of a Discrete Random Variable' The standard deviation of a discrete random variable measures the spread of the variable's probability distribution. The higher the value, the more the different values of the variable are spread apart from one another. The equation for the standard deviation of a discrete random variable is: σ = √(Σx2P(x)-μ2), where: σ = standard deviation Example: σ = √(Σx2P(x)-μ2) Insert summation values: σ = √(2.6-(1.32)) Square the mean: σ = √(2.6-1.69) Subtract: σ = √(.91) Calculate square root: σ = .9539... 'Factorials' n!, read as the factorial of n, is the product of n and every whole number leading up to n, or: n! = 1 x 2 x 3 x 4 ... (n - 2) x (n - 1) x n Example: 5! = 1 x 2 x 3 x 4 x 5 5! = 120 'Combinations' A combination is a collection of elements taken from a larger set of elements, where different orders of elements ARE NOT counted seperately. Example: Combinations of the letters A, B, C, and D (A, B), (A, C), (A, D), (B, C), (B, D), (C, D) Note that order does not matter; (A, B) and (B, A) are considered to be the same combination and are not counted seperately. The formula for combinations is nCx, where: n = number of elements available x = number of elements to be selected The formula for finding the number of combinations is: nCx = n!/(x!(n-x)!) Example: 5C3 = n!/(x!(n-x)!) Insert values of variables 5C3 = 5!/(3!(5-3)!) Subtract 5C3 = 5!/(3!2!) Simplify factorials 5C3 = 5*4*3!/3!2! 5C3 = 5*4/2*1 5C3 = 5*2 5C3 = 10 There are 10 different combinations when selecting 3 elements from a set of 5. 'Permutations' A permutation is a collection of elements taken from a larger set of elements, where the different orders of elements ARE counted seperately. Example: Permutations of the letters A, B, and C (A, B), (A, C), (B, A), (B, C), (C, A), (C, B) The permutations (A, B) and (B, A) are bother counted individually because their elements are in different orders. The formula for permutations is nPx, where: n = number of elements available x = number of elements to be selected The formula for finding the number of permutations is: nPx = n!/(n-x)! Example: 5P3 = n!/(n-x)! Insert values of variables 5P3 = 5!/(5-3)! Insert values of variables 5P3 = 5!/(5-3)! Subtract 5P3 = 5!/2! Simplify factorials 5P3 = 5*4*3*2!/2! 5P3 = 5*4*3 5P3 = 60 There are 60 different permutations when selecting 3 elements from a set of 5. 'Binomial Probability Distribution' The binomial probability distribution is the equation used to determine the odds that a certain number of tests are successful, after a certain number of tests have been performed. The equation for finding the odds of a certain number of tests are successful is P(x) = (nCx)(p^x)(q^x-n), where: P(x) = odds that x number of tests are successful n = number of tests performed p = probability of success q = probability of failure (1 - p) x = number of successes n - x = number of failures Example: A coin is tossed 5 times. Assuming a 50% chance of winning, find the chance of winning 3 tosses out of 5. P(x) = (nCx)(p^x)(q^x-n) Insert values of variables P(x) = (5C3)(.5^3)(.5^5-3) Simplify P(x) = (10)(.125)(.5^2) P(x) = (10)(.125)(.25) P(x) = (10)(.125)(.25) P(x) = .3125 The odds of winning 3 coin tosses out of 5 are .3125, or 31.25% See the graphing calculators tutorial to learn how to use the "binompdf(" function to perform this equation Mean and Standard Deviation of the Binomial Distribution While the formulas for finding the mean and standard deviation of discrete random variables are still applicable, simpler formulas are available for binomial distributions: μ = np σ = √(npq), where: n = number of tests p = probability of success q = probability of failure 'Hypergeometric Probability Distribution' Hypergeometric Probability Distribution is similar to binomial probability distribution, with the exeption that the outcome of a test will affect the probabilities of the outcomes of later tests. The equation for hypergeometric probability distribution is: P(x) = (rCx)(N-rCn-x)/(NCn), where N = number of elements in a population r = number of successes in a population N - r = number of failures in a population n = number of trials in a sample x = number of successful trials n - x = number of failed trials It can also be read as: P(x) = (popwinsCsamplewins)(poplossesCsamplelosses)/(popsizeCsamplesize), or P(x) = (WINS C wins)(LOSSES C losses)/(TOTAL C total) For example: There is a bag of 10 marbles. 3 marbles are blue. If 5 marbles are randomly chosen, find the probability that 2 of those chosen marbles are blue. P(x) = (rCx)(N-rCn-x)/(NCn) Insert values of variables P(x) = (3C2)(10-3C5-2)/(10C5) Simplify P(x) = (3C2)(7C3)/(10C5) P(x) = (3)(35)/(252) P(x) = 105/(252) P(x) = 0.4166666667 The probabilty of 2 of the chosen marbles being blue is 41.67% 'Poisson Probability Distribution' The Poisson probability distribution is one process used to find the chance of an event occuring a given number of times withing a given interval, under the assumption that the event is a discrete random variable, and that the outcomes are both random and independent of one another. The equation for the Poisson probability distribution is: P(x) = (λ^x)(e^-λ)/x!, where x = given number of occurances of an event λ = mean number of occurances e = approximately 2.71828 For example: You own a small clothing store. On average, you sell 5 shirts per day. What are the odds that you sell exactly 3 shirts on any given day? P(x) = (λ^x)(e^-λ)/x! P(x = 3) = (5^3)(2.71828^-5)/3! Simplify P(x = 3) = (125)(.0067379697)/3*2 P(x = 3) = .8422462076/6 P(x = 3) = .1403743679 There is a 14.04% chance that you will sell exactly 3 shirts in a single day. To learn how to enter the Poisson probability distribution into a graphing calculator, refer to the Operating a TI-83+ Graphing Calculator Tip: Inputting this formula using Microsoft Excel In column A, label the variables λ to A1, x to A2, and e to A3 In column B, insert the values of the different variables in B1, B2, and B3 Underneath the values, insert the following formula to find the probability: =(B1^B2)*(B3^-B1)/FACT(B2) Adjust the targeted cells as required